Electronic Integration

An electronic integrator is a device that produces an output waveform whose value at any instant of time equals the total area under the input waveform up to that point in time. (For those familiar with mathematical integration, the process produces the time-varying function I:, Vi,,(/) dt.) To illustrate this concept, suppose the input to an electronic integrator is the de level E volts, which is first connected to the integrator at an instant of time we will call t = O. Refer to Figure 14-18. The plot of the de “waveform” versus time is simply a horizontal line at level E volts, since the de voltage is constant. The more time that we allow to pass, the greater the area that accumulates under the dc waveform. At any time-point t, the total area under the input waveform between time 0 and time 1 is (height) X (width) = EI volts, as illustrated in the figure. For example, if E = 5 V de, then the output will be 5 V at t = 1 s, 10 V at t = 2 s, 15 V at t = 3 s, and so forth. We see that the output is the ramp voltage vo(t) = Et.

When the input to a practical integrator is a de level, the output will rise linearly with time, as shown in Figure 14-18, and will eventually reach the maximum possible output voltage of the amplifier. Of course, the integration process ceases at that time. If the input voltage goes negative for a certain interval of time, the total area during that interval is negative and subtracts from whatever positive area had previously accumulated, thus reducing the output voltage. Therefore, the input must periodically go positive ane} negative to prevent the output of an integrator from reaching its positive or negative limit. We will explore this process in greater detail in a later discussion on waveshaping. .

How an electronic integrator is constructed using an operational amplifier. Note that the component in the feedback path is capacitor C, and that the amplifier is operated in an inverting configuration. Besides the usual idealamplifier assumptions, we arc assuming zero input offset, since any input dc level would be integrated as shown in Figure 14-18 and would eventually cause the amplifier to saturate. Thus, we show an ideal-integra/or circuit.

sing the standard symbol (/I to represent integration of the voltage v between time 0 and time t, we an show that, the output of this circuit is

This equation shows that the output is the (inverted) integral of the input, multiplied by the constant 11R, C. If this circuit were used to integrate the dc waveform shown in Figure 14-18, the output would be a negative-going ramp (u, = – EtIR, C). Readers unfamiliar with calculus can skip the present paragraph, in which we demonstrate why the circuit of Figure 14-19 performs integration. Since the current into the – input is 0, we. have, from Kirchhoff’s cu~rent law, where i, is the input current through R, and iI’ is the feedback current through the capacitor. Since v = 0, the current in the capacitor is

Hereafter, we will use the abbreviated symbol I to represent integration. It can be shown. using calculus, that the mathematical integral of the sine wave A since. The most important fact revealed by equation 14-46 is that the output of an integrator with sinusoidal input is a sinusoidal waveform whose amplitude is inversely proportional to its frequency. This observation follows from the presence of w (= 21T f) in the denominator of (14-46). For example, if a IOO-Hz input sine wave produces an output with peak value 10 Y, then, all else being equal, a 200- Hz sine wave will produce an output with peak value 5 Y. Note also that the output leads the input by 90°, regardless of frequency, since cos wt = sin(wt + 90°). However, as shown by equation 14-46, this phase relation is due to both integration and the inverting action of the amplifier. In many practical applications, noninverting integrator circuits are used to introduce phase lag into  a system:

1. Find the peak value of the output of the ideal integrator shown in Figure 14-20. The input is Vi. = 0.5 sin (lOOt) Y.
2. Repeat, when Vin = 0.5 sin ( V.

A voltage-controlled current source that can be operated with a grounded load. To understand its behavior as a current source, refer to  which shows the voltages and currents in the circuit. Since there is (ideally) Zero current into the + input, Kirchhoff’s current law at the node where R,. is connected to the + input gives

Solution. From equation 14-31, lc = Vilt/R = (10 V)/(41dl) = 2.5 mA. Therefore, the voltage at node C (Vd is Vc = hRL = (2.5 mA)(1.5 kn) = 3.75 V. We know that the voltage at node B is twice Vc (Vo = 2Vd: VB = 2Vc = 2(3.75) = 7.5 V. ‘The voltage at node A is one-half that at node B (u: = Vol2): VA = (O(VB) = 0)(7.5) = 3.75 V. The currents II> Iz, 13, and 14 in RI, Rz, R3, and R4 can then be found:

Posted on November 19, 2015 in Applications of Operational Amplifiers

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